Question: Find the complex number $z$ such that
\[|z - 1| = |z + 3| = |z - i|.\]
Let $z = a + bi,$ where $a$ and $b$ are real numbers.  Then
\[|(a - 1) + bi| = |(a + 3) + bi| = |a + (b - 1)i|.\]Hence, $(a - 1)^2 + b^2 = (a + 3)^2 + b^2 = a^2 + (b - 1)^2.$

From $(a - 1)^2 + b^2 = (a + 3)^2 + b^2,$ $8a = -8,$ so $a = -1.$  Then the equations above become
\[4 + b^2 = 1 + (b - 1)^2.\]Solving, we find $b = -1.$  Therefore, $z = \boxed{-1 - i}.$